Pumps for the FE Civil Exam: Pump Power, System Curves, NPSH and Affinity Laws

NCEES FE Civil spec: Water Resources and Environmental Engineering, C. Pumps. FE Reference Handbook: Fluid Mechanics chapter (pump power, NPSH, pump curves, affinity laws).

What the FE Civil Spec Covers

Section C is pumps. Civil exam questions focus on what a pump must do in a water or wastewater system: the total dynamic head it adds, the power it draws, where it operates on its curve, whether it will cavitate, and how its performance changes with speed or with a second pump. Most problems are a short energy balance followed by a power calculation.

Key Concepts

Total dynamic head (TDH): static lift plus friction and minor losses plus any change in pressure head and velocity head between the two points.
System curve: head required versus flow, $H = H_{s t a t i c} + k Q^{2}$ . The operating point is where it crosses the pump curve.
Efficiency: water power divided by shaft (brake) power. Motor efficiency, if given, applies on top of pump efficiency.
NPSH: absolute head at the suction above vapor pressure. NPSH available must exceed NPSH required.
Affinity laws: for the same pump at a new speed, flow scales with N, head with N squared and power with N cubed.

Core Formulas

TDH = (z_{2} - z_{1}) + \frac{p _{2} - p _{1}}{γ} + \frac{V _{2}^{2} - V _{1}^{2}}{2 g} + h_{L}

P_{w a t er} = γ Q H P_{b r ak e} = \frac{γ Q H}{η _{p u m p}}

In US units, a handy form is horsepower = Q(gpm) times H(ft) divided by 3,960, for water.

NPSH_{A} = \frac{p _{a t m} - p _{v a p or}}{γ} \pm z_{s} - h_{L, s u c t i o n}

Use plus z when the supply surface is above the pump and minus z when the pump lifts from below.

\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}} \frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2} \frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}

Where It Is in the FE Reference Handbook

The pump power equation, NPSH available, a sample centrifugal pump curve and the scaling (affinity) laws for fans, pumps and compressors are all in the Fluid Mechanics chapter. Search "Pump Power", "NPSH" or "Affinity". The handbook version of the affinity laws also includes impeller diameter D; for a fixed impeller, D drops out and only the speed ratio remains.

Worked Example

A pump lifts 0.08 m³/s of water from a sump to a tank whose water surface is 25 m higher. Both surfaces are open to the atmosphere and the pipe losses total 5 m. Pump efficiency is 75%. (a) Find the brake power. (b) If the pump speed is reduced from 1,750 rpm to 1,450 rpm, estimate the new flow, head and power on the scaled pump curve.

Step 1, TDH. Pressure and velocity heads are zero at both open surfaces:

H = 25 + 5 = 30 m

Step 2, power:

P_{w a t er} = (9.81 kN/m^{3}) (0.08 m^{3} / s) (30 m) = 23.5 kW

P_{b r ak e} = \frac{23.5}{0.75} = 31.4 kW

Step 3, affinity laws with speed ratio 1,450/1,750 = 0.829:

Q_{2} = 0.08 (0.829) = 0.066 m^{3} / s H_{2} = 30 (0.829)^{2} = 20.6 m

P_{2} = 31.4 (0.829)^{3} = 17.9 kW

Answer: about 31.4 kW at full speed. Note that 20.6 m is less than the 25 m static lift, so at 1,450 rpm this pump could no longer deliver water to the tank. The affinity laws move a point along the pump curve; the real operating point is still found where the new pump curve meets the system curve.

How FE Questions Are Framed

A typical pump question gives two reservoir elevations, a pipe loss (or the data to compute it with Darcy-Weisbach or Hazen-Williams), a flow and an efficiency, and asks for power. Work in a consistent unit system: in SI, gamma in kN/m³ times Q in m³/s times H in m gives kW directly. In US units, gamma of 62.4 lb/ft³ times Q in cfs times H in ft gives ft-lb/s, and 550 ft-lb/s is one horsepower.

Operating point questions give a pump curve as a table or graph and a system curve equation. Evaluate the system head at two or three flows and find where the two curves meet. For NPSH questions, read the vapor pressure for the stated temperature from the water property table and keep every pressure term absolute. If a question asks what happens when a valve is throttled, the system curve gets steeper, flow drops and the pump head rises along its curve.

Common Mistakes

Multiplying by efficiency instead of dividing when going from water power to brake power.
Leaving out friction losses or the velocity head at a free discharge when building TDH.
Using gauge pressure in NPSH. Atmospheric and vapor pressure must be absolute.
Squaring the speed ratio for flow or cubing it for head. Flow is linear, head is squared, power is cubed.
Adding heads for parallel pumps. Parallel pumps add flow, series pumps add head.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A pump moves water at Q = 0.05 m³/s through a total head of 20 m with 80% pump efficiency. What is the required shaft input power? Use ρ = 1,000 kg/m³ and g = 9.81 m/s². See the worked solution or practice the Fluids & Hydraulics deck
A pump moves water between two large reservoirs open to the atmosphere. The static elevation rise is 20 m and total pipe losses are 5 m. What pump head is required? See the worked solution or practice the Fluids & Hydraulics deck
A pump draws 20 °C water from a reservoir whose surface is 3.0 m below the pump centreline. Friction loss in the suction line is 1.2 m. Atmospheric pressure is 101.3 kPa, the vapour pressure of water at 20 °C is 2.34 kPa, and the specific weight of water is 9.79 kN/m³. The net positive suction head available is most nearly: See the worked solution or practice the Fluids & Hydraulics deck

Related Calculators and Examples

Frequently Asked Questions

What is the difference between water power and brake power?

Water (hydraulic) power is the power delivered to the fluid, gamma times Q times total head. Brake power is the shaft power the motor must supply, which is water power divided by pump efficiency, so it is always larger.

How do pumps in series and in parallel combine?

Identical pumps in series add their heads at the same flow. Identical pumps in parallel add their flows at the same head. The actual operating point is still where the combined pump curve crosses the system curve.

When does a pump cavitate?

Cavitation risk exists when the net positive suction head available at the pump inlet is less than the net positive suction head required by the manufacturer. Raising the supply level, shortening the suction line or lowering the pump all increase NPSH available.

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